Let $G$ be the mapping class group of a closed surface $S_{g}$. Bestvina-Bromberg-Fujuwara http://front.math.ucdavis.edu/1006.1939 recently constructed a finite index subgroup $B$ of $G$ such that for every essential closed simple curve $\gamma$ on $S_{g}$ and every $h\in B$ either $h\gamma=\gamma$ or $h\gamma$ intersects $\gamma$ (everything is modulo isotopy of $S_{g}$). The construction is not difficult: $B$ is just the subgroup of $G$ fixing $\pi_1(S_{g})/N$ for some characteristic subgroup $N$ of the $\pi_1$ of finite index. The $N$ can be found by intersecting kernels of the first homology mod $6$ with kernels of the first homology mod 2 of all index 2 subgroups of the $\pi_1$. Question: can one find another finite index subgroup $B'$ of $G$ with the same property but with a smaller index. Most probably $B'$ cannot be above Torelli subgroup, but can it be "not far from Torelli". I am interested in $B'$ that does not act non-trivially on a simplicial tree. The motivation is here: http://front.math.ucdavis.edu/1005.5056.

Update: I read somewhere (I do not remember where now) that if $g$ is an element of the Torelli subgroup, $\gamma$ is a simple closed curve, then at least two of the three curves $\gamma, g\gamma, g^2\gamma$ must intersect. From this I deduced that two curves are not enough and that a Bestvina-Bromberg-Fujiwara subgroup cannot contain the Torelli subgroup. Now the question is whether it may contain the Johnson kernel. For this one needs first to answer the following question: if $g$ is the Dehn twist about a separating curve, $\gamma$ is a simple closed curve, is it true that $g\gamma$ intersects $\gamma$ or $g\gamma=\gamma$?

3 Answers
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Regarding the new question, you're correct that $f\in \mathcal{I}$ doesn't imply $f(\gamma)$ meets $\gamma$: for example, if $f$ is a product of disjoint bounding pairs and $\gamma$ is a (nonseparating) curve meeting each curve-defining-$f$ in at most 1 point, then $f\in\mathcal{I}$ but $f(\gamma)$ is disjoint from $\gamma$.

For the rest of the question, the following theorems are taken from Farb-Leininger-Margalit, "The lower central series and pseudo-Anosov dilatations", American Jour. Math, 130 (2008), no. 3, 799--827. (PDF)

Lemma 2.2: if $f\in \mathcal{I}$, $\gamma$ is separating, and $f(\gamma)\neq\gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$.

Lemma 2.3: if $f\in \mathcal{I}$ and $\gamma$ is nonseparating, and $f(\gamma)\neq\gamma$ then $i\big(f(\gamma),\gamma\big)\geq 2$ or $i\big(f^2(\gamma),\gamma\big)\geq 2$.

Proposition 3.2: if $f\in\mathcal{K}$ and $f(\gamma)\neq \gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$.

In particular 3.2 answers your question about a separating twist. The proof of the proposition is as follows: $\mathcal{K}$ contains no nonseparating bounding pairs (use the Johnson homomorphism). By Lemma 2.2 we can assume $\gamma$ is nonseparating. But if $f(\gamma)$ is disjoint frmo $\gamma$, then $[f,T_{\gamma}]=f T_{\gamma}f^{-1} T_{\gamma}^{-1}=T_{f(\gamma)}T_{\gamma}^{-1}$ is a bounding pair, which lies in $\mathcal{K}$ if $f$ does, yielding a contradiction.

I do appreciate the nice calming green background, but I'm not sure we're done. If $f\in\mathcal{K}$, and $g\in \mathcal{BBF}$ (the group defined by Bestvina-Bromberg-Fujiwara), how do you know that $fg$ has the desired property?

Here is a sketch of a possible construction of a BBF-subgroup which properly contains the Johnson kernel. Let $G$ be the subgroup generated by the Johnson kernel $\mathcal{K}$ together with ($\mathcal{BBF}\cap\mathcal{I}$). Since $G$ is contained in $\mathcal{I}$, separating curves are taken care of by Lemma 2.2 above. If $\gamma$ is a nonseparating curve such that $f(\gamma)$ is disjoint from $\gamma$, then $[f,T_\gamma]$ is a nonseparating bounding pair; $[f,T_\gamma]$ lies in $G$ if $f$ does, since $G$ is normal.

Now the idea is to show that $G$ contains no nonseparating bounding pairs. Certainly $\mathcal{BBF}$ doesn't, but we need to show that this is still true after throwing in $\mathcal{K}$. Since $G$ contains $\mathcal{K}=\ker \big(\tau\colon\mathcal{I}\to \bigwedge^3 H/H\big)$, this is the same as showing that $\tau(\mathcal{BBF}\cap \mathcal{I})$ does not contain the image under $\tau$ of a genus $k$ bounding pair. We know the image of such a BP, and you should be able to write down a condition on $\tau(f)$ for $f\in\mathcal{BBF}$ coming from the triviality of the action on the homology of the double cover. I have the feeling that this condition will rule out such a bounding pair, which would conclude the proof. I don't know how hard this approach would be if you wanted to actually carry it out, though.

@Tom: Thanks! So in principle a BBF subgroup can contain the Johnson kernel.
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Mark SapirOct 7 '10 at 13:48

In fact it just occurred to me that if we take the Johnson kernel and the BBF subgroup constructed in the paper of Bestvina-Bromberg-Fujiwara, and take the subgroup generated by the two, we get a BBF subgroup which is bigger than the Johnson kernel. Right? So I will accept Tom's answer.
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Mark SapirOct 7 '10 at 13:59

@Tom: Yes, you are right, I was too quick accepting your answer. But it is justified by your addition which does look promising. Perhaps instead of BBF, you can take a finite index subgroup of it. We cannot intersect BBF with the Torelli subgroup because we want to obtain a finite index subgroup of $G$. The resulting subgroup should a) be of finite index and b) contain the Johnson kernel $K$.
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Mark SapirOct 8 '10 at 2:04

I don't have an answer to Mark's question but perhaps I can make a few comments. I will work in $A= Aut$ instead of $G=MCG$ since it makes the discussion a little simpler.

What Ian says is correct even in the punctured case. Two disjoint curves can be distinguished by their ${\mathbb Z}_2$-homology in double covers. In particular if $\alpha$ and $\beta$ are disjoint then there is some double cover where either both curves have lifts representing different homology classes or one curve has a lift and the other doesn't.

Using this the group $B$ can be chosen to be the finite index subgroup of $A$ that acts trivially on the ${\mathbb Z_2}$-homology of the union of all double covers. However, as Mark says we wanted the stronger property that for any subsurface $Y$ either $gY = Y$ or $\partial gY$ intersects $\partial Y$. If $Y$ is a sphere with three boundary components and $S_{g,p}$ has low complexity then it could be that $Y$ and $gY$ are complementary so if you are concerned about thrice punctured sphere you need to go to a smaller subgroup. For our applications and I would guess any application you can probably ignore thrice punctured spheres. In the posted version of the paper we weren't very efficient about the construction of the group but hopefully we'll soon post a new version with this more streamlined version.

I'm not sure how much bigger one can make $B$ but here is an idea. Let $H$ be the ${\mathbb Z_2}$-homology of the union of double covers and let $A'$ be the automorphisms of $H$. The group $A$ maps to the finite group $A'$ and we can rephrase the previous paragraph by saying that any subgroup $B'$ of $A$ that maps to the trivial subgroup of $A'$ has the property that if $h \in B'$ than for any curve $\gamma$ either $\gamma = h\gamma$ or $\gamma$ and $h\gamma$ intersect. I think it should be possible to find a non-trivial subgroup $K$ of $A'$ such that if the image of $B'$ is contained in $K$ the same property holds.

Even if one could find a bigger group $K$ it is not clear if this would give you bigger subgroups of $A$ because it could be that any subgroup $B'$ whose image was in $K$ would actually have trivial image in $A'$. For example if the image of Torelli or the Johnson kernel in $A'$ was the same as the image of $A$ then this wouldn't help at all.

@Ken: It is too late here already, so I am a bit slow. Do you think that a subgroup with the same property as $B$ but containing the Johnson kernel can be constructed or not?
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Mark SapirOct 7 '10 at 3:10

Dear Ken and others: regarding the action of the Torelli group on the homology of double covers, check out Looijenga, "Prym representations of mapping class groups", Geom. Ded. 64 (1997) 1, 69-83, ams.org/mathscinet-getitem?mr=1432535. Your description looks to be the intersection of the kernels of the order-2 Prym representations. Looijenga proves that the image of a Prym representation is arithmetic. In the last paragraph of the introduction he seems to imply that the image of Torelli is the same (or perhaps finite index), though I can't find where in the paper he proves this.
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Tom ChurchOct 7 '10 at 4:05

@Mark: I don't know if such a $B$ can be constructed and don't have a guess either way. I was only suggesting a way to push the construction a bit further. For my suggestion to be useful you would need to now something about the action of the Johnson kernel (or Torelli) on the homology of double covers. In particular at minimum it would have to be smaller than the action of the whole mapping class group. Maybe the paper Tom mentioned gives some information about this.
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Ken BrombergOct 7 '10 at 4:53

@Ian, Tom and Ken: If $g$ is the Dehn twist about a separating curve, $\gamma$ is another simple closed curve, is it true that either $g\gamma=\gamma$ or $g\gamma$ intersects $\gamma$?
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Mark SapirOct 7 '10 at 5:43

@Ken: I'm copying the comment I made to my own answer (because I wasn't sure you would see it). If an (orientable) mapping class acts trivially on mod 4 homology, then it cannot map a surface to its complement, since it must preserve the orientation of the boundary curves. Since the second derived 2-subgroup contains the kernel of the map to $H_1(S_g,\mathbb{Z}/4)$, I still think it suffices for your question. (welcome to MO!)
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Ian AgolOct 7 '10 at 16:21

For $S_g$, I think you only need kernels of the first homology $\mod 2$ of all index $2$ subgroups of $\pi_1(S_g)$ (I haven't thought about the punctured case yet). Call this subgroup $K$. Suppose $h$ acts trivially on $\pi_1(S_g)/K$. First of all, notice that $h$ acts trivially on $H_1(S_g,\mathbb{Z}/2)$ (all further coefficients will be understood to be in $\mathbb{Z}/2$). If $\gamma$ is separating in $S_g$, then it divides $S_g$ into subsurfaces $A$ and $B$, say with $\chi(A)>\chi(B)$. Then since $h\gamma\cap \gamma=\emptyset$, we have $h(A)\cap A=\emptyset$, which means that $h(H_1(A))\neq H_1(A)$, a contradiction.

Now suppose that $\gamma$ is non-separating. Then since $h$ acts trivially on $H_1(S_g)$, $[\gamma]=[h(\gamma)]\in H_1(S_g)$. So $\gamma\cup h\gamma$ divides $S_g$ into subsurfaces $A$ and $B$. Consider the 2-fold cover $\tilde{S_g}\to S_g$ dual to $\gamma$, so that the preimage of $\gamma$ consists of two curves, $\tilde{\gamma}$ which separate $\tilde{S_g}$. Let $\tilde{h}$ be the lift of $h$ to $\tilde{S_g}$ which acts trivially on $H_1(\tilde{S_g})$ (which exists since by hypothesis $h$ acts trivially on $\pi_1(S_g)/K$). Then $\tilde{\gamma}\cup \tilde{h}\tilde{\gamma}$ divides $\tilde{S_g}$ into regions $\tilde{A_1},\tilde{A_2}$ and $\tilde{B_1},\tilde{B_2}$ which are each trivial covers of $A$ and $B$, respectively (labeled so that $\tilde{A_i}\cup \tilde{B_i}$ bounds $\tilde{\gamma}$, $i=1,2$). Then one sees that $\tilde{h}(\tilde{A_1}\cup\tilde{B_1})=\tilde{A_1}\cup \tilde{B_2}$, possibly up to relabeling. Then $\tilde{h}(H_1(\tilde{B_1}))\not=H_1(\tilde{B_1})$, which contradicts that $\tilde{h}$ acts trivially on $H_1(\tilde{S_g})$ (since it acts trivially on $\pi_1(S_g)/K$).

I'm not sure if an element of the Torelli group can have this property.

Ian: thanks! There are some misprints in the answer which make it difficult to understand (at least for me). Say, "non-separating" --> "separating"? What does the last phrase mean? If I understood Mladen Bestvina's explanations correctly, kernel of mod 3 homology was needed to use orientation of the boundary components of the subsurface (distinguish two complementary subsurfaces). You can avoid that? Also your $B$ is still too small. I wonder if there exists a $B$ with a really small index and possibly above the Johnson kernel which does the same
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Mark SapirOct 6 '10 at 18:01

Mark: I think I fixed the misprints. The orientation isn't important for this argument. Actually, one could keep track of the orientations in principle because $h$ acts trivially on $H_1(S_g,\mathbb{Z}/4)$, since the kernel of this is contained in $K$. The last phrase means that I'm not sure if there's an element $h$ of the Torelli group and a non-separating curve $\gamma$ such that $\gamma\cap h\gamma=\emptyset$.
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Ian AgolOct 6 '10 at 20:23

@Ian: thanks again. Still the last phrase in your answer looks mysterious. Certainly $B$ intersects the Torelly subgroup (and the intersection is of finite index in the Torelli subgroup). Do you think such a subgroup $B$ can contain the Johnson kernel (the subgroup generated by the Dehn twists of separating curves)? Being non-normal is OK.
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Mark SapirOct 6 '10 at 20:56

Mark, the last phrase just refers to your question of whether B' could be chosen to contain Torelli. If there were an element of Torelli which took a curve disjoint from itself, then B' couldn't contain Torelli.
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Ian AgolOct 6 '10 at 21:59